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The function \( p \) is defined as \( p(x)=\left\{\begin{array}{ll}2 x+1 & \text { if } x>1 \\ -x^{2}+1 & \text { if }-1
Ask by Bowman Fuentes. in South Africa
Feb 26,2025

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Answer

The domain of function \( p \) is \( (-1, 1) \cup (1, \infty) \), which is choice (E).

Solution

The function p is defined piecewise with one definition for x > 1 and another for -1 < x < 1. There are no definitions provided for x ≤ -1 or x = 1. Thus, the values for which p is defined are in the interval (-1, 1) (not including -1 and 1) and the interval (1, ∞). In other words, the domain of p is: (-1, 1) ∪ (1, ∞) Therefore, the correct answer is (E).

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The correct choice for the domain of function \( p \) is (E) \( D p=(-1,1) \cup(1, \infty) \). This is because the function has two parts: for \( x > 1 \) (where \( p(x) = 2x + 1 \)) and for \( -1 < x < 1 \) (where \( p(x) = -x^2 + 1 \)). The function is not defined for \( x \leq -1 \) and \( x = 1 \), which leads to the specified domain being a combination of the two intervals. Now let's dive a bit deeper: The piecewise nature of functions like \( p(x) \) often has interesting historical origins in mathematics. The development of piecewise functions evolved through calculus and mathematical analysis, where mathematicians sought ways to describe different behaviors in a single expression—significantly aiding in the study of real-world phenomena such as physics and engineering systems, where systems can change behavior under different conditions. When working with piecewise functions, one of the common mistakes is overlooking the endpoints or boundaries—like \( x = -1 \) and \( x = 1 \) in this case. It's essential to check these boundary points and how they relate to the definition of the function, as they can change the outcome significantly. Always ensure you evaluate whether the function is continuous across its intervals and that all potential input values are considered!

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